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Find all functions defined on the set of positive reals which take positive real values and satisfy: $$ f(xf(y))=yf(x) $$ for all ; $ f(x)\to0 $ and as $ x\to\infty $

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This has nothing to do with functional analysis. –  kahen Oct 26 '12 at 2:14

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Let $f(x) = f(y)$. Then $x f(x) = f(xf(x)) = f(xf(y)) = yf(x) \implies x = y$.

Substitute $x=1$, and $y=1$ separately and compare to get $f(f(f(x))) = f(x)$. Hence $f(f(x)) = x$. Now, substitute $x=y=1$ to get $f(f(1)) = f(1)$ and hence $f(1) = 1$.

Put $g(x) = xf(x)$. Then we have that $f(g(x)) = g(x)$.

Now, if $f(z) = z$, then substituting $x = \frac1z, y= z$ gives $f(\frac1z) = \frac1z$. So if $g(x)$ is not identically equal to $1$, then we have some $z>1$ for which $f(z) = z$. Substitute $x=y=z$ to get $f(z^2) = z^2$. By induction, $f(z^{2^n}) = z^{2^n}$. This is not possible, since the RHS goes to $\infty$, while the $LHS$ goes to 0 by the given condition as $n \to \infty$. Thus $g(x) =1$ identically, and hence $f(x) = \frac1x$. Substituting back, we see that this satisfies the given equation, so this is the unique solution.

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Any constant function satisfies $f(f(f(x)))=f(x)$. –  N. S. Oct 26 '12 at 6:13
    
@N.S. Ouch. Edited. –  ronno Oct 26 '12 at 6:20

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